Problem:
Show that $$\ln2 = \sum_\limits{n=1}^\infty\frac1{n2^n}.$$
My progress:
The problem before this one had me find the Taylor series for $\ln(1-x)$ which was $$-\sum\limits_{n=1}^\infty \frac{x^n}n$$
so I figured I'd use $x=-1$ and plug that into the Taylor series. However, there was a side note stating that the Taylor series I found is only valid for $x\in(-1, 1)$. And in any case, my calculation isn't going anywhere, since I end up with the series $1-\frac12+\frac13-\frac14\cdots$
Question 1: How can I use this to solve the problem stated initially, and in the title?
Question 2: I can see why $x$ is restricted to be less than 1, to prevent taking the log of zero or a negative. Why is it not valid for x greater than 1?
$\endgroup$ 104 Answers
$\begingroup$Hint: $\ln 2 = - \ln (1/2)$. Can you use that and your result on the series for $\ln (1-x)$ to solve the problem?
$\endgroup$ 1 $\begingroup$A different perspective: Your series is the Euler transform of $$ 1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\ldots$$ and so it has an equal sum.
$\endgroup$ 2 $\begingroup$Hint:
$$\frac1{1-x}=\sum_{n=0}^\infty x^n\;,\;\;|x|<1\implies-\log(1-x)=\sum_{n=0}^\infty\frac{x^{n+1}}{n+1}=x+\sum_{n=2}^\infty\frac{x^n}n$$
$\endgroup$ $\begingroup$For similar cases, where you cannot simply evaluate inside the radius of convergence... There is Abel's theorem with conditions that let you evaluate exactly on the radius of convergence.
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